Problem: Recently Duff has been a soldier in the army. Malek is her commander.

Their country, Andarz Gu has n cities (numbered from 1 to n) and n - 1 bidirectional roads. Each road connects two different cities. There exist a unique path between any two cities.

There are also m people living in Andarz Gu (numbered from 1 to m). Each person has and ID number. ID number of i - th person is i and he/she lives in city number ci. Note that there may be more than one person in a city, also there may be no people living in the city.

<image>

Malek loves to order. That's why he asks Duff to answer to q queries. In each query, he gives her numbers v, u and a.

To answer a query:

Assume there are x people living in the cities lying on the path from city v to city u. Assume these people's IDs are p1, p2, ..., px in increasing order. 

If k = min(x, a), then Duff should tell Malek numbers k, p1, p2, ..., pk in this order. In the other words, Malek wants to know a minimums on that path (or less, if there are less than a people).

Duff is very busy at the moment, so she asked you to help her and answer the queries.

Input

The first line of input contains three integers, n, m and q (1 ≤ n, m, q ≤ 105).

The next n - 1 lines contain the roads. Each line contains two integers v and u, endpoints of a road (1 ≤ v, u ≤ n, v ≠ u).

Next line contains m integers c1, c2, ..., cm separated by spaces (1 ≤ ci ≤ n for each 1 ≤ i ≤ m).

Next q lines contain the queries. Each of them contains three integers, v, u and a (1 ≤ v, u ≤ n and 1 ≤ a ≤ 10).

Output

For each query, print numbers k, p1, p2, ..., pk separated by spaces in one line.

Examples

Input

5 4 5
1 3
1 2
1 4
4 5
2 1 4 3
4 5 6
1 5 2
5 5 10
2 3 3
5 3 1


Output

1 3
2 2 3
0
3 1 2 4
1 2

Note

Graph of Andarz Gu in the sample case is as follows (ID of people in each city are written next to them):

<image>

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
const int N = 1e5 + 10, L = 20, MOD = 1e9 + 7, MAXN = 1e5 + 10;
const int inf = 2e9 + 10;
vector<int> v[N], cnt[N][20], id[N];
int pr[N][20], h[N];
void dfs(int u, int par) {
  pr[u][0] = par;
  cnt[u][0] = id[u];
  for (int i = 1; i <= 17; i++) {
    if ((1ll << i) > h[u]) {
      continue;
    }
    pr[u][i] = pr[pr[u][i - 1]][i - 1];
    cnt[u][i] = cnt[pr[u][i - 1]][i - 1];
    for (int j = 0; j < (int)cnt[u][i - 1].size(); j++) {
      cnt[u][i].push_back(cnt[u][i - 1][j]);
    }
    sort(cnt[u][i].begin(), cnt[u][i].end());
    if (cnt[u][i].size() > 10) cnt[u][i].resize(10);
  }
  for (int i = 0; i < (int)v[u].size(); i++) {
    if (v[u][i] != par) {
      h[v[u][i]] = h[u] + 1;
      dfs(v[u][i], u);
    }
  }
}
inline int lca(int x, int y) {
  if (h[x] < h[y]) {
    swap(x, y);
  };
  int dif = h[x] - h[y];
  for (int i = 0; i <= 17; i++) {
    if ((dif & (1ll << i)) > 0) {
      x = pr[x][i];
    }
  }
  if (x == y) {
    return y;
  };
  for (int i = 17; i >= 0; i--) {
    if (pr[x][i] != pr[y][i]) {
      x = pr[x][i];
      y = pr[y][i];
    }
  }
  return pr[x][0];
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int n, m, q;
  cin >> n >> m >> q;
  for (int i = 1; i <= n - 1; i++) {
    int x, y;
    cin >> x >> y;
    v[x].push_back(y);
    v[y].push_back(x);
  }
  for (int i = 1; i <= m; i++) {
    int x;
    cin >> x;
    id[x].push_back(i);
  }
  for (int i = 1; i <= n; i++) {
    sort(id[i].begin(), id[i].end());
    if (id[i].size() > 10) id[i].resize(10);
  }
  dfs(1, 0);
  for (int t = 1; t <= q; t++) {
    int x, y, a;
    cin >> x >> y >> a;
    if (h[x] < h[y]) {
      swap(x, y);
    };
    vector<int> ret;
    int p = lca(x, y);
    int difx = h[x] - h[p], dify = h[y] - h[p];
    for (int i = 0; i <= 17; i++) {
      if ((difx & (1ll << i)) > 0) {
        for (int j = 0; j < (int)cnt[x][i].size(); j++) {
          ret.push_back(cnt[x][i][j]);
        }
        x = pr[x][i];
      }
    }
    for (int i = 0; i <= 17; i++) {
      if ((dify & (1ll << i)) > 0) {
        for (int j = 0; j < (int)cnt[y][i].size(); j++) {
          ret.push_back(cnt[y][i][j]);
        }
        y = pr[y][i];
      }
    }
    for (int i = 0; i < min((int)id[p].size(), a); i++) {
      ret.push_back(id[p][i]);
    }
    sort(ret.begin(), ret.end());
    int cnt = min((int)ret.size(), a);
    cout << cnt << " ";
    for (int i = 0; i < cnt; i++) {
      cout << ret[i] << " ";
    }
    cout << "\n";
  }
  return 0;
}